A problematic mismatch in the total number of rigid surface operators between the \(B_n\)  and the \(C_n\)  theories was pointed out in [1]} [2]}. In [3]}, we clear up cause of this discrepancy and construct all the \(B_n/C_n\)  rigid surface operators which can not have a dual. The fingerprint invariant is assumed to be equivalent to the symbol invariant [2]}.
In [5]}, we prove the symbol invariant of partitions implies the fingerprint invariant of partitions. The constructions of the symbol invariant and the fingerprint invariant in previous works play a significant role in the proof.
This problem suggest us to find map between two invariants.
